Does $\mathsf{ZFC} + \neg\mathrm{Con}(\mathsf{ZFC})$ suffice as a foundations of mathematics?

Question

I've heard people make the argument that:

$\mathsf{ZFC}$ suffices as a foundations of mathematics because almost all theorems in the mathematics literature can be proven using $\mathsf{ZFC}$, so long as they're suitably translated into the language of sets.

Now let $\mathrm{Con}(\mathsf{ZFC})$ denote the statement that $\mathsf{ZFC}$ is consistent, formalized in the language of $\mathsf{ZFC}$. By Godel's second incompleteness theorem, if $\mathsf{ZFC}$ is consistent, then so too is $\mathsf{ZFC}+\neg\mathrm{Con}(\mathsf{ZFC})$.

Observe also that the above argument works for $\mathsf{ZFC}+\neg\mathrm{Con}(\mathsf{ZFC})$.

Thus, would $\mathsf{ZFC}+\neg\mathrm{Con}(\mathsf{ZFC})$ suffice as a foundations? And if so, why is nobody using it? (okay, clearly there are psychological reasons why nobody is using it - but are there deeper reasons?)

@HaraldHanche-Olsen, yes, $\mathrm{Con}(\mathsf{ZFC})$ is the statement that $\mathsf{ZFC}$ is consistent, formalized in the language of $\mathsf{ZFC}$. — goblin GONE, Apr 23 '13 at 10:59
Well, the extra axiom seems quite useless for doing mathematics, does it not? — Harald Hanche-Olsen, Apr 23 '13 at 11:03
@HaraldHanche-Olsen, I don't know! Perhaps it simplifies certain proofs. — goblin GONE, Apr 23 '13 at 11:17
If ZFC is inconsistent, then this theory is inconsistent too. If ZFC is consistent then this theory has the wrong natural numbers (it has natural numbers that don't exist). I guess that's something undesirable. — Apostolos, Apr 23 '13 at 11:18
@HaraldHanche-Olsen It's not entirely useless. It can be used to prove a whole range of negative results, like the non-existence of large cardinals! — Zhen Lin, Apr 23 '13 at 11:20

score 10 · Accepted Answer · edited Apr 23 '13 at 22:06

10

There are three main problems with the theory you propose.

Often when we want to use forcing we assume there are countable transitive models of $\sf ZF$ (with additional axioms, of course, such as choice, large cardinals, and so on).

Indeed this problem is merely cosmetic, we can work around it using several approaches, but it would require us to say "Let $M$ be a countable transitive model of enough axioms of $\sf ZF$" every time, which is quite annoying.
A more severe problem would be with large cardinals. If $\sf ZFC$ is internally inconsistent, then there are no large cardinals in the universe, and there are no set models of $\sf ZFC$ with large cardinals. Since those are needed for more than a handful results, and they do make a wonderful and deep research topic in set theory, the assumption that $\sf ZFC$ is internally inconsistent would simply eliminate this entire field.
Often we like to think of the integers in the meta-theory (i.e. in the logic outside the universe) as the same integers as those of the universe. But in a universe where $\sf ZFC$ is internally inconsistent there is a proof of that, but this proof cannot be encoded by a standard integer. Therefore there must be a non-standard integer.

As with the issue about forcing, this too is probably a cosmetic and convenience issue to some extent, but it is also a psychological point that I find reassuring to think that the "real" universe of sets have the same integers as those we can write in the first-order logic of its meta-theory.

From the last point, let me draw my final conclusion, which is probably a psychological issue here. Foundational theories should be something that you believe is a good basis to develop mathematics within. If you believe that $\sf ZFC$ is a reasonable foundation, then of course you can talk about models of $\sf ZFC+\lnot\rm Con(\sf ZFC)$ because it is a larger theory. But it would mean that from a foundational point of view, you probably don't think that $\sf ZFC$ is consistent, so your entire foundation is inconsistent.

The fact that $\sf ZFC+\lnot\rm Con(\sf ZFC)$ is consistent, if $\sf ZFC+\rm Con(\sf ZFC)$ is consistent, is a peculiarity of how mathematics works. Much like the many other peculiarities that we can find in universes of set theory.

edited Apr 23 '13 at 22:06

goblin GONE

69,755

answered Apr 23 '13 at 13:46

Asaf Karagila

408,303

"but it is also a psychological point that I find reassuring to think that the 'real' universe of sets have the same integers as those we can write in the first-order logic of its meta-theory" -- why is that? If you assume a 'real' universe of sets, why does it still make sense to talk about a metatheory with a possibly different set of integers? Why should such a metatheory be reasonable? Another explanation could be that you do believe in ZFC, but you do not believe Con(ZFC) really expresses what it is supposed to express. – Jori Apr 27 '20 at 11:59
Not critique, but just curious to understand what are reasonable positions an why (also from a philosophical point of view). Of course you are right that this is not a very good theory because of the other points you mentioned, in particular the loss of the study of large cardinals and results that depend on them. – Jori Apr 27 '20 at 12:01
But Con(ZFC) is a very robust thing, and we know that if you work in ZFC, then this is equivalent to the existence of a model. Just because you disagree with the coding doesn't mean that it invalidates the result. The point is that we don't want to think about the universe of mathematics as having a different notion of integers than what we do. It's a sort of absurd approach to mathematics. – Asaf Karagila Apr 27 '20 at 12:12
Maybe I understand your thought process: by agreeing that ZFC + ¬Con(ZFC) is a reasonable foundation of mathematics you implicitly agree that the 'real' universe of sets models this ('models' as in the sense that many mathematicians think of the cumulative hierarchy as the 'real' model of ZFC, not 'models' in a formal sense). But then there are two options: if you think that ZFC is inconsistent (meta-theoretically) then you must also think that this new theory is inconsistent. Otherwise, if you think ZFC is consistent (meta-theoretically) then you must conclude that the 'real' universe has ... – Jori Apr 27 '20 at 12:28
non-standard integers, which is pretty weird if it's supposed to be the 'real' universe. However I'm worried that this necessary step into the metatheory is unjustified or incoherent: if you assume a 'real' universe, why would there be anything beyond that described by a metatheory? I think that is what is confusing me primarily. – Jori Apr 27 '20 at 12:31
@Jori: That's the idea, yes. – Asaf Karagila Apr 27 '20 at 12:44
But what does a metatheory refer to if we have already decided on a 'real' universe? – Jori Apr 27 '20 at 14:10
It's where we formalise our proofs about mathematics. – Asaf Karagila Apr 27 '20 at 14:11

score 9 · Answer 2 · answered Apr 23 '13 at 13:29

The problem with the system you propose is that we add to ZFC a statement that in the background we assume (or hope) that is false. Specifically, doing mathematics using ZFC relies on the fact that (a sizeable fragment of) ZFC is consistent. We don't want to have a contradictory system, because contradictions are analytic falsehoods that never take place and (hence) imply everything (at least in the most common logical systems).

The statement $\mathrm{Con(ZFC)}$ is a statement about finite objects (e.g. numbers) and the fact that we have a lot more intuition about these kinds of objects implies that we have a lot more expectations about the accuracy (with respect to our intuition) of a theory that describes these objects.

So, if ZFC is inconsistent so is the theory that you propose (since it extends ZFC). On the other hand, if ZFC is consistent, any model of the theory that you propose will have numbers (or finite objects) that should not exist. The witness of the existential statement that you propose we add will not be an actual natural number (or something that our intuition will describe as a natural number).

Personally, I think this is a strong argument as to why this system is not a good extension of ZFC. We want set theory to accurately describe finite objects, and the extra axiom that you propose goes against that.

Does $\mathsf{ZFC} + \neg\mathrm{Con}(\mathsf{ZFC})$ suffice as a foundations of mathematics?

2 Answers2